AP Calc AB (Serrano) Chapter 5 Notes
Find d/dx = ∫(x²)(-2x) sin t dt
*lower limit needs to be a constant so set value a between then so -2x < a < x² = ∫(a)(-2x) sin t dt + ∫(x²)(a) sin t dt = -∫(-2x)(a) sin t dt + ∫(x²)(a) sin t dt = -sin (-2x)(-2) + sin(x²)(2x) = 2 sin (-2x) + 2x sin (x²)
Find the derivative of the function d/dx ∫(x⁴)(x) sec t dt
*use Chain Rule x⁴ = u; d/dx = du/dx * d/du d/dx ∫(u)(0) sec t dt = sec u = sec u * du/dx = sec (x⁴) * 4x³
Find the approximate total distance traveled given the following (time, velocity): (0,25) (5,31) (10,35) (15,43) (20,47) (25,46) (30,41) - Use LRAM (the velocity at the beginning of each time interval) - Use RRAM (the velocity at the end of each time interval)
- 5(25 + 31 + 35 + 43 + 47 + 46) = 1135, stop at second to last - 5(31 + 35 + 43 + 47 + 46 + 41) = 1215 ft 1135 < x < 1215
Find the following examples of indefinite integrals: - ∫(10x⁴ + 3x³)dx - ∫ (x⁻¹)dx
- = 2x⁵ + (3/4)x⁴ + C - ∫ (1/x)dx = ln |x| + C, where x is an absolute value because the domain of y = ln x is always x > 0
What are the three major kinds of RAM?
- RRAM (right rectangular approximation method) - LRAM (left rectangular approximation method) - MRAM (midpoint rectangular approximation method)
What are the two situations where it's impossible to find the exact value of a definite integral?
- it's difficult or impossible to find the antiderivative - there is no formula for the function such as data collected in the experiment
Describe the following on graphs for definite integrals: - area above the x axis - area below the x axis
- positive; add for net area - negative; subtract for net area
Describe how we evaluate definite integrals, or the exact area under f(x) from [a,b] using the following: - geometrically - algebraically
- the entire area up to b minus the entire area up to a will equal just the area between a and b *(b, -∞) - (a, -∞) - F(b) - F(a) where the area of each is represented by the antiderivative
How can this be represented graphically?
- the x axis is time (4 s) and the y axis is velocity (3 ft/s) - velocity is a straight line because there's no acceleration - the distance that the object travels is equal to the area under the line
You are given a graphed function f whose domain is [-3,5]. Let g(x) = ∫(x)(2) f(t)dt. 1) Find g(3) 2) Find the instantaneous rate of change of g at x = 4
1) find g(-3) so ∫(-3)(2) f(t)dt = -∫(2)(-3) f(t)dt = - [(.5*3*-3) + (.5*1*3) + (.5*1*3)] = 9/2 2) g'(4) = f(4) = 3
You are given a graphed function f whose domain is [-3,5]. Let g(x) = ∫(x)(2) f(t)dt. 3) Write the equation for the line tangent to the graph of g at x = 4 4) At what value does g(x) have an absolute min or max? State what they are
3) g'(4) = 3; (4, g(4)) g(4) = ∫(4)(2) f(t)dt = 9 y = 9 - 3(x - 4) 4) endpoints and critical values of g(x) so where g' = 0 (at x = 1) g(1) = ∫(1)(2) = -∫(2)(1) f(t)dt = -(3/2) g(-3) = 9/2 g(5) = ∫(5)(2) f(t)dt = 5(4) + ∫(5)(4) f(t)dt = 9 + 1/2(1*3) = 9 3/2 so there is a max at 5 and a min at 1
Evaluate ∫(3)(1) e^x dx
= e^x](3)(1) = e³ - e
Evaluate ∫(π/2)(0) cos θ dθ
= sin θ](π/2)(0) = F(π/2) - F(0) = 1 - 0 = 1
Integrate the following indefinite integral: ∫tan x dx
= ∫ sin x/cos x *must test sin x and cos x to find which one works for u and will cancel; u must = cos x because if u = sin x ∫ = sin x/-cos² x dx = -du/sin x - ∫ 1/u du = - ln |cos u| + c
According to the FTC where f is continuous on [a,b] and ∫(b)(a) f(x)dx = F(b) - F(a), where F is any antiderivative of f, what does F' equal?
F' = f
f(t) is the function defined on the interval [-5,5] by a graph below and the area function F is defined by F(x) = ∫(x)(-5) f(t)dt. a) Where is F increasing and decreasing? b) where is the absolute maximum and absolute minimum of F? c) Where are the local extreme points? d) Where is F CU and CD? e) Where are the inflection points of F?
F' = f a) increasing when +, decreasing when - b) check at endpoints and where f(x) crosses graph: absolute max at F(3.5) and min at F(5) c) where f = 0 d) CU when increasing, CD when decreasing e) inflection points when changes from increasing to decreasing
What is the definition of a definite integral?
If f is a continuous function defined for a < x < b, we divide the interval [a,b] into n subintervals of equal width ∆x = (b-a)/n. We let x₀ (=a) x₁, x₂, . . . xₙ (=b) be the endpoints of these subintervals and we let x₁*, x₂*,. . .xₙ* be any sample points in these subintervals, so xi* lies in the ith subinterval [x(i-1), x(i)]. Then the definite integral from a to b is: ∫(b)(a) f(x)dx = lim(n→∞) ⁿε(i = 1) f(xi*) ∆x
Evaluation Theorem
If f is continuous on the interval [a,b] then ∫(a)(b) f(x)dx = F(b) - F(a) where F is any antiderivative of f, that is F' = f
Let f(x) = 0 if x > 0, x if 0 ≤ x ≤ 1, 2 - x if 1 < x ≤ 2, and 0 if x > 2. g(x) = ∫(x)(0) f(t)dt. a) Find an expression for g(x) similar to the one for f(x)
If x > 0, g(x) = ∫(x)(0) f(t)dt = ∫(x)(0) 0dt = 0 If 0 ≤ x ≤ 1, g(x) = ∫(x)(0) t dt = ∫(x)(0) 1/2x²](x)(0) = 1/2x² If 1 < x ≤ 2, g(x) = ∫(x)(0) f(t)dt = ∫(1)(0) f(t)dt + ∫(x)(1) f(t)dt = g(1) + ∫(x)(1) (2-t)dt = 1/2(1)² + (2t - 1/2t²)](x)(1) = 1/2 + (2x -1/2x²) - (2 - 1/2) = -1/2x² + 2x + 1 If x > 2, then ∫(x)(0) = g(2) + (x)(2) 0dt = 0 + 1 = 1 So g(x) = 0 if x < 0, 1/2x² if 0 ≤ x ≤ 1, -1/2x² + 2x + 1 if 1 < x ≤ 2, and 1 if x > 2
For 0 ≤ t ≤ 6, a particle is moving along the x-axis. The particle's position, x(t), is not explicitly given. The velocity of the particle is given by v(t) = 2 sin(e^(t/4)). The acceleration of the particle is given by a(t) = 1/2e^(t/4) cos (e^(t/4)) and x(0) = 2. c) Find the total distance traveled by the particle from t = 0 to t = 6.
Math -> Num -> Abs() to get absolute value plug in ∫(6)(0) (|v(t)|) dx where v(t) is the given velocity function = 12.573 m
What was the approximated area under the y = x² from [0,1] using Reimann sums? Try again using the Evaluation Theorem
M₄ = 0.34 Evaluation Theorem: ∫(1)(0) x² dx = x³/3](1)(0) = F(1) - F(0) = 1/3 - 0 = 0.33
What is the difference between Riemann sums and definite intervals?
Riemann sums are approximate because they use RAM; εf(xi*)∆x has height as the first component and width as the second Definite integrals give the exact area: the new formula is ∫(a)(b) f(x)dx and it is the area under the curve y = f(x) from a to b. f(x) is height and dx is width
Why does g'(x) = f(x)?
When you derive g in g(x) = ∫(x)(a) f(x)dt, then you derive the integral and they cancel each other out
How can a definite integral be interpreted?
a net area or a difference of sums: ∫(a)(b) f(x)dx = A₁ - A₂ where A₁ is above the x axis and A₂ is below
You are given a graphed function f whose domain is [-3,5]. Let g(x) = ∫(x)(2) f(t)dt. Interpret: a) g(x) b) g'(x) c) concavity of g is g"(x) =
a) area under f from x = 2 b) a point on the function f c) the rate of change/slope of f (aka g')
Let g(x) = ∫(x)(0) f(t)dt, where f is the function whose graph is shown. a) evaluate g(0), g(1), g(6), etc. b) on what interval is g increasing? c) where does g have a maximum value? d) sketch a rough graph of g
a) find the area under each from 0 to each limit/number b) the part of the graph that is positive since g'(x) = f(x) c) find the area of where g(x) = 0 and the max and mins d) graph each of the areas for each point of g(x)
Use Part 1 of the FTC to find the derivative of the function: a) g(y) = ∫(y)(2) sin t dt b) h(x) = ∫(x²)(0) √(1 + r³)dr
a) g'(y) = y² sin t dt b) u = x² and du/dx = 2x = √(1 + u³)2x = 2x√(1 + x⁶)
A particle moves along a line so that its velocity at time t is v(t) = t² - t - 6 m/s a) Find the displacement of the particle during the time period 1 ≤ t ≤ 4 b) Find the distance traveled during this time period
a) net area = displacement: ∫(4)(1) (t² - t - 6)dx (t³/3) - (t²/2) - 6t](4)(1) = F(4) - F(1) = -9/2 m b) t² - t - 6 = (t-3)(t+2) = 3, -2 = 3 only because within ∫(4)(1) |v(t)|dx = ∫(3)(1) |v(t)|dx + ∫(4)(3) |v(t)|dx (t³/3) - (t²/2) - 6t](3)(1) = |(27/3 - 9/2 - 18) - (1/8 - 1/2 - 6)| (t³/3) - (t²/2) - 6t](4)(3) = |(64/3 - 16/2 - 24) - (27/3 - 9/2 - 18)| = 61/6 m
For 0 ≤ t ≤ 6, a particle is moving along the x-axis. The particle's position, x(t), is not explicitly given. The velocity of the particle is given by v(t) = 2 sin(e^(t/4)). The acceleration of the particle is given by a(t) = 1/2e^(t/4) cos (e^(t/4)) and x(0) = 2. a) Is the speed of the particle increasing or decreasing at time t = 5.5? Give a reason for your answer.
a) plug into calculator: v(5.5) = -0.45337 a(5.5) = -1.35851 the speed is increasing at t = 5.5 s because the velocity and acceleration are the same sign
Answer the following: a) If w'(t) is the rate of growth of a child in pounds per year, what does ∫(10)(5) w'(t)dt represent? b) If oil leaks from a tank at a rate of r(t) gallons per minute at time t, what does ∫(120)(0) r(t)dt represent?
a) ∫(10)(5) w'(t)dt represents the increase in the child's weight between 5 and 10 b) ∫(120)(0) r(t)dt represents the number of gallons leaked in 120 minutes
How should you solve problems asking you to interpret an integral?
acknowledge bounds and units use change and increase/decrease if ALWAYS increasing or decreasing
What does an indefinite integral ∫ f(x)dx represent?
an entire family of antiderivatives
What is the difference between an indefinite and definite integral?
an indefinite integral has no bounds and a definite integral does
For 0 ≤ t ≤ 6, a particle is moving along the x-axis. The particle's position, x(t), is not explicitly given. The velocity of the particle is given by v(t) = 2 sin(e^(t/4)). The acceleration of the particle is given by a(t) = 1/2e^(t/4) cos (e^(t/4)) and x(0) = 2. b) Find the average velocity of the particle for the time period 0 ≤ t ≤ 6
average velocity = ∆x/∆t plug in ∫(6)(0) (v(t)) dx where v(t) is the given velocity function = 11.696 11.696 m/(6-0 s) = 1.949 m/s
Why doesn't which RAM method you use matter for integrals?
because it involves infinite numbers (the limit approaches infinity)
Why are both parts called the Fundamental Theorem of Calculus?
because they relate to integration and differentiation
What must a *definite* integral have?
bounds
Integrate the following indefinite integral: ∫sin⁴ x cos x dx
can be interpreted as (sin x)⁴ cos x dx so let u = sin x du/dx = - cos x so dx = -du/cos x ∫ u⁴ cos x * du/cos x = ∫u⁴ du antiderive and plug in u: u⁵/5 + c = sin⁵ u/5 + c
What are Riemann sums?
different forms of RAM (MRAM, LRAM, RRAM, etc.)
Given an object moving at a constant rate of 3 ft/s, how far has the object gone after 4 seconds?
distance = rate * time 3 ft/s * 4 s = 12 ft
Find ∫(9)(1) [(2t² + t²√t - 1)/t²]dt
divide each component individually by t² = (2 + √t - 1/t²)dt = 2t + (2/3)t³/² + t⁻¹](9)(1) = F(9) - F(1) (18 + 18 + 1/9) - (2 + 2/3 + 1) = 32 4/9
How do you find the area under any function when velocity is not constant?
divide it into lines of equal width
How do you integrate variants of e^x?
ex. ∫e^-3x dx u = -3x = -e^-3x/3 ex. ∫u^6 > ∫u^7/7
What is the midpoint rule with definite integrals?
find the change in x and use midpoints ∆x = b-a/n
Evaluate the following integral by interpreting it in terms of geometric area: ∫(3)(0) (x-1)dx
from 0-3, the graph has a small shaded triangle under the x axis (above the function) and a larger shaded triangle above the x axis (below the function) A = bh/2 so = (1)(1)/2 = -1/2 (2)(2)/2 = 2 2 - 1/2 = 1.5
If f is continuous on [a,b] and g(x) = ∫(x)(a) f(t)dt, then what is g'(x) equal to according to the Fundamental Theorem of Calculus?
g'(x) = f(x)
Find the derivative of the function g(x) = ∫(x)(0) √(1 + t²) dt
g'(x) = √(1 + x²)
If g(x) = ∫(x)(a) f(t)dt where a = 1 and f(t) = t², find the formula for g(x) and calculate g'(x)
g(x) = ∫(x)(1) t²dt antiderive t³/3](x)(1) = x³/3 - 1/3 g'(x) = d/dx(1/3x³ - 1/3) = x²
Describe g, or the "area so far" function
g(x) is the area and the function is y = f(t)
How would you find the graph of g for the above problem?
graph x then g(x): (0,0), (1,1), (2,3), (3,4.3), (4,3), (5, 1.7) then let the undefined part of the graph gradually slope off to the side
What are we learning in this chapter?
how to find the exact area under a curve using the equation of the curve
Describe indefinite integrals
if F is an antiderivative of f on an interval, then the most general antiderivative f on I is F(x) + C, where C is an arbitrary constant **indefinite integrals = antiderivative + C
What can g(x) be interpreted as if f happens to be a positive function?
it is the area under the graph of f from a to x, where x can vary from a to b
Integrate the following indefinite integral: ∫√(1 + x²)*2x dx
let u = 1 + x² ∫(√u)(2x)(dx) du/dx = 2x; dx = du/2x ∫(√u)(2x)(dx) * du/2x = ∫(√u)U du or u¹/² du antiderive then plug in u: (2/3)u³/² + c = [2(1+x²)³/²]/3
Integrate the following indefinite integral: ∫√(4x - 1)dx
let u = 4x - 1 ∫√(u) dx; du/dx = 4 so dx = du/4 = ∫√(u) * du/4 antiderive and plug in u: (2/3)u³/² du/4 + c = (4x-1)³/²/6 + c
Integrate the following definite integral: ∫(π/2)(0) cos³ x sin x dx
let u = cos x; dx = du/-sin x ∫u³ sin x * du/-sin x = -∫u³ du u(π/2) = cos π/2 = 0 u(0) = cos 0 = 1 - ∫(0)(1) u³ du = ∫(1)(0) u³ du antiderive: u⁴/4](1)(0) = 1/4 - 0 = 1/4
Integrate the following indefinite integral: ∫ x² sin(x³)dx
let u = x³ du = 3x²dx so dx = du/3x² ∫x² sin u * du/3x² = ∫sin u du/3 antiderive and plug in u: = 1/3 cos(x³) + c
Integrate the following definite integral: ∫(1)(-1) 3x² √(x³+1) dx
let u = x³ + 1 du/dx = 3x²; dx = du/3x² ∫(3x²)√(u) * du/3x²; ∫ = √u du u(1) = 2, u(-1) = 0 ∫(2)(0) (2/3)u³/²](2)(0) = 2√8/3
Consider the area under the curve y = x² from [0,1]. Use RRAM with four rectangles under the curve
n = # of rectangles ∆x = b-a/n = (1-0)/4 so each = 1/4 *place a point on the curve at 1/4, 1/2, 3/4, and 1. Draw a line across the width of the section and lines down to the x axis as needed R₄ = 1/4f(1/4) + 1/4f(1/2) + 1/4f(3/4) + 1/4f(1) 1/4[f(1/4) + f(1/2) + f(3/4) + f(1)] 1/4[(1/4)² + (1/2)² + (3/4)² + (1)²] = 0.46875
Use the Midpoint Rule with n = 5 to approximate ∫(2)(1) 1/x(dx)
n = 5; approximate ∫(2)(1) 1/x(dx) ∆x = (2-1)/5 = 1/5 ∫(2)(1) 1/x(dx) ≈ 1/5[f≈(1.1) + f(1.3) + f(1.5) + f(1.7) + f(1.9)] = 1/5(1/1.1 + 1/1.3 + 1/1.5 + 1/1.7 + 1/1.9) = 0.691908
Do definite integrals use + C?
no
What happens as we use more rectangles (n)?
our area approximation becomes better
Let g(x) = ∫(x)(0) f(t)dt on a given graph. Find g(0), g(1), g(2), g(3), g(4), and g(5)
pay attention to details: the lower bound is 0 and the upper bound (v) is unknown, so find using geometry g(0) = ∫(0)(0) f(t)dt = 0 g(1) = ∫(1)(0) f(t)dt = bh/2 = 1*2/2) = 1 + 0 = 1 g(2) = ∫(2)(1) f(t)dt = bh = 1*2 = 2 + 1 = 3 g(3) = g(2) + ∫(3)(2) f(t)dt = 1.3 (approximate triangle) = 3 + 1.3 = 4.3 g(4) = g(3) + ∫(4)(3) f(t)dt = 4.3 - 1.3 = 3 g(5) = g(4) + ∫(5)(4) f(t)dt = 3 - 1.3 = 1.7
For 0 ≤ t ≤ 6, a particle is moving along the x-axis. The particle's position, x(t), is not explicitly given. The velocity of the particle is given by v(t) = 2 sin(e^(t/4)). The acceleration of the particle is given by a(t) = 1/2e^(t/4) cos (e^(t/4)) and x(0) = 2. d) For 0 ≤ t ≤ 6, the particle changes direction exactly once. Find the position of the particle at that time.
plug in y = v(t) and find where it crosses the x axis Calc -> 2(zero) A = 5.19552 when v(t) = 0 x(A) = x(0) + ∫(A)(0) v(t)dt 2 + ∫(A)(0) y1 dx = 14.134
You are given a graphed function f whose domain is [-3,5]. Let g(x) = ∫(x)(2) f(t)dt. 5) At what x-values does g(x) have a point of inflection?
point of inflection when g"(x) changes signs so when f'(x) changes signs (increasing to decreasing) x =0,3
Evaluate the following integral by interpreting it in terms of geometric area: ∫(1)(0) √(1-x²)(dx)
r = 1 because y = ±√(r² - x²) the given function is positive so only consider the hemisphere above the x axis, and only consider the graph in quadrant I (a quarter of the circle) because the function is from 0-1 A = πr² and r = 1 and the area is a quarter of the circle so A = πr²/4 = π/4
Find ∫(0)(-3) (1 + √(9 - x²))dx geometrically using area
r = 3 1/4(π)(3)² + 1(3)
How can we find the area under the curve without knowing the equation of the curve?
rectangular approximation methods (RAM)
What should you do if the upper bound isn't on top?
reverse and make the bound negative ex. ∫(-2)(0) = -∫(0)(-2)
Integrate the following indefinite integral: ∫(e^x + 1)/e^x
simplify to ∫1 + e^-x u = -x; e^u * -1du = x + -∫e^u du = x - e^x + c
What can you use to rewrite functions in a form that you can integrate?
substitution or change of variable
What is a definite integral?
the area under a curve from point a to b (think of it as displacement)
What is the derivative when the lower limit is a constant and the upper limit is x for g(x) = ∫(x)(a) f(t)dt?
the derivative is f with respect to x, or f(x)
What is wrong with ∫(3)(-1) 1/x²dx = x⁻¹/-1](3)(-1) = -4/3?
the function is not continuous because the bounds include 0 and 1/0² = DNE
What properties does a definite integral have because f is continuous?
the limit in this definition always exists (not DNE) and gives the same value no matter how we choose the sample points xi*
What is the definite integral defined as?
the limit of Riemann sums
What should you substitute for u?
the most interior function **NEVER use x
Why is MRAM sometimes better?
to avoid over or underestimation, MRAM is sometimes better especially for increasing or decreasing functions
Integrate the following definite integral: f is continuous on ∫(4)(0) f(x)dx = 10. Find ∫(2)(0) f(2x)dx
u = 2x; dx = 1/2 du 1/2 ∫(4)(0) f(u) du = 1/2[10] = 5
Integrate the following indefinite integral: ∫(sin x)/(1 + cos x) dx
u = cos x; dx = du/-sin x = ∫(sin x)/(1 + u²)dx * -du/sin x = -∫1/(1 + u²) du = -tan⁻¹ u = -tan⁻¹ (cos x) + C
Integrate the following definite integral: ∫(e^4)(e) 1/(x√ln x)) dx
u = ln x; dx = x du u(e^4) = 4; u(e) = 1 ∫(4)(1) 1/x u¹/² I x du = ∫(4)(1) u⁻¹/² du [2u¹/² du](4)(1) = 4 - 2 = 2
Integrate the following definite integral: ∫(1/2)(0) sin⁻¹ x/√(1 - x²) dx
u = sin⁻¹ x; du = 1/√(1-x²) dx; dx = √(1-x²)du u(1/2) = π/6; u(0) = 0 ∫(π/6)(0) u du = 1/2u²](π/6)(0) = π²/72
Integrate the following definite integral: ∫(2)(1) x√(x-1)
u = x-1; du = 1 dx u(1) = 0, u(2) = 1 x - 1 = u; x = u + 1 ∫(1)(0) x√u du ∫(1)(0) (u+1)u¹/² ∫(1)(0) (u³/² + u¹/²)du = u⁵/² = 2/5u⁵/² + 2/3u³/²](1)(0) = 2/5 + 2/3 - (0+0) = 16/15
Integrate the following definite integral: ∫(√π)(0) x cos (x²) dx
u = x²; dx = du/2x u(0) = 0; u(√π) = π ∫(π)(0) x cos u * 1/2x du 1/2 ∫(π)(0) cos u du = 1/2 [sin u](π)(0) = 1/2(0-0) = 0
Find d/dx ∫(5x)(2) 3t² dt
use chain 3(5x)² * 5 = 375x²
What does the area under the curve g depend on?
what x is (not t!), where x is a bound as represented by g(x) = ∫(x)(a) f(t)dt
When is RRAM an overestimate?
when the function is always increasing the right estimate is over
What is the function of a circle?
x² + y² = r²; y = ±√(r² - x²) the positive part is the hemisphere above the x axis and the negative part is the hemisphere below
Find the interval on which the curve y = ∫(x)(0) = 1/(1 + t + t²) dt is concave upward
y' = 1/(1 + x + x²)dx y" = (1+x+x²)⁻¹*dx = -1/(1+x+x²)² * (1 + 2x) 1+ x + x² is always positive so 1 + 2x > 0 x < -1/2 or (-∞, -1/2)
How does integrating definite integrals differ from integrating indefinite integrals?
you must reevaluate the bounds
What should you do when calculating RAM algebraically?
you need equal intervals and must find different rectangles with widths if this is not provided
Consider the area under the curve y = x² from [0,1]. Use LRAM with four rectangles under the curve
∆x = b-a/n = 1-0/4 so each = 1/4 * place a point on the curve at 0, 1/4, 1/2, and 3/4. Draw a line right across the width of the rectangle then down so it connects with the x axis. The leftmost rectangle at f(0) collapses. L₄ = 1/4[f(0) + f(1/4) + f(1/2) + f(3/4) = 1/4(0 + 1/16 + 1/4 + 9/16) = 0.21875 so 0.21875 < A < 0.46875
Consider the area under the curve y = x² from [0,1]. Use MRAM with four rectangles under the curve
∆x = b-a/n = 1-0/4 so each = 1/4 *place a point on the curve at the midpoints of each rectangle (1/8, 3/8, 5/8, 7/8) and draw a line across the width of the rectangle before drawing a line down to connect it with the x axis M₄ = 1/4[f(1/8) + f(3/8) + f(5/8) + f(7/8)] = 0.32421875
The velocity of a car was read from its speedometer at 10-second intervals and recorded in the table. Use the Midpoint Rule to estimate the distance traveled by the car. From left to right: (time in seconds, v (mi/hr)) (0,0), (10,38), (20,52), (30, 58), (40, 55), (50, 51), (60, 56), (70, 53), (80, 50), (90, 47), (100, 45)
∫(100)(0) v(t) dt using the midpoint rule for 0 ≤ t ≤ 100 with n = 5, the length of each of the five time intervals is 20 seconds. 20/3600 hr = 1/180 1/180[v(10) + v(30) + v(50) + v(70) + v(90)] = 247/180 or 1.4 miles
Use Part 1 of the FTC to find the derivative of the function: g(x) = ∫(3x)(2x) (u²-1)/(u²+1) du
∫(3x)(2x) f(u)du = ∫(0)(2x) f(u)du + ∫(3x)(0) f(u)du = -∫(0)(2x) f(u)du + ∫(3x)(0) f(u)du = -((2x)² - 1)/(2x² +1))*2x + ((3x²)-1)/(3x²)+1)) * 3x = -2(4x²-1)/(4x²+1) + 3(9x²-1)/(9x²+1)
If it is known that ∫(10)(0) f(x)dx = 17 and ∫(8)(0) f(x)dx = 12, find ∫(10)(8) f(x)dx
∫(8)(0) f(x)dx +∫(10)(8) f(x)dx = ∫(10)(0) f(x)dx 12 + ∫(10)(8) f(x)dx = 17 ∫(10)(8) f(x)dx = 5
What is the addition property of an integral?
∫(b)(a) [f(x) + g(x)]dx = ∫(b)(a) f(x)dx + ∫(b)(a) g(x)dx
What is the subtraction property of an integral?
∫(b)(a) [f(x) - g(x)]dx = ∫(b)(a) f(x)dx - ∫(b)(a) g(x)dx
What is the constant property of an integral?
∫(b)(a) c dx = c(b-a) where c is any constant
What is the coefficient property of an integral where c is any constant?
∫(b)(a) cf(x) dx = c ∫(b)(a) f(x)dx